(x-4)(x-4)=2x^2-5x-12

Solution for (x-4)(x-4)=2x^2-5x-12 equation:

(x-4)(x-4)=2x^2-5x-12

We move all terms to the left:

(x-4)(x-4)-(2x^2-5x-12)=0

We get rid of parentheses

-2x^2+(x-4)(x-4)+5x+12=0

We multiply parentheses ..

-2x^2+(+x^2-4x-4x+16)+5x+12=0

We get rid of parentheses

-2x^2+x^2-4x-4x+5x+16+12=0

We add all the numbers together, and all the variables

-1x^2-3x+28=0

a = -1; b = -3; c = +28;
Δ = b2-4ac
Δ = -32-4·(-1)·28
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*-1}=\frac{14}{-2}
=-7
$